Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

class Main {


    public static double prob(int N, double[] proba) {

        double[][] dp = new double[N + 1][N + 1];


        dp[0][0] = 1;

        for(int coin = 1; coin <= N; coin++) {
            dp[coin][0] = dp[coin - 1][0] * (1- proba[coin-1]);
        }

        for(int coin = 1; coin <= N; coin++) {
            dp[coin][coin] = dp[coin-1][coin-1] * (proba[coin-1]);
        }

        for(int coin=1; coin<=N; coin++) {
            for(int heads = 1; heads <= coin; heads++) {
                dp[coin][heads] = dp[coin - 1][heads - 1] * proba[coin - 1] + dp[coin - 1][heads] * (1 - proba[coin - 1]);
            }
        }

        double totalProba = 0.0;

        for(int heads=N/2+1; heads<=N; heads++) {
            totalProba = totalProba + dp[N][heads];
        }
        return totalProba;
    }


    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();

        double[] p = new double[N];

        for(int i=0; i<N; i++)
            p[i] = scanner.nextDouble();

        System.out.println(prob(N, p));

    }
}